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Red ribbon string is put around two circular disks as shown above. If

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Bunuel
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Red ribbon string is put around two circular disks as shown above. If [#permalink] Red ribbon string is put around two circular disks as shown above. If   05 May 2020, 10:48
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Red ribbon string is put around two circular disks as shown above. If the radii of the two disks are 40 cm and 30 cm, respectively, what is the total length of the string?


A. \(140\) cm

B. \(140 + \frac{165π}{2}\) cm

C. \(140 + 105π\) cm

D. \(140 + 120π\) cm

E. \(140 + 165π\) cm

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C
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Re: Red ribbon string is put around two circular disks as shown above. If [#permalink] Red ribbon string is put around two circular disks as shown above. If   05 May 2020, 11:08
Bunuel wrote:

Red ribbon string is put around two circular disk as shown above. If the radii of the two disk are 40 cm and 30 cm, respectively, what is the total length of the string?


A. \(140\) cm

B. \(140 + \frac{165π}{2}\) cm

C. \(140 + 105π\) cm

D. \(140 + 120π\) cm

E. \(140 + 165π\) cm

Show Spoiler
Attachment:
Untitled.png


Given: Red ribbon string is put around two circular disk as shown above.

Asked: If the radii of the two disk are 40 cm and 30 cm, respectively, what is the total length of the string?

Total length of string = 270/360 * 2π *(30 + 40) + 2*(40+30) = 70 (3π/2 + 2) = 140 + 105π

IMO C
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Re: Red ribbon string is put around two circular disks as shown above. If [#permalink] Red ribbon string is put around two circular disks as shown above. If   06 May 2020, 03:11
*********************************

So, Since the ribbon covers 3/4th of the ribbon, use the formula:

3/4( 2 Pi R)

So, you would have 3/4(2*Pi* 40) = 3/4(2*Pi*30)
which will equal to 105 Pi. -----1
Keeping that aside, let's work on the intersecting tangential lines.
we know that three angles in the respective quadrilaterals are 90. so, the fourth angle also should be 90.

which makes all the sides equal(as the radius of a circle is equal at any point on the circumference).

for tangent 1, we will get (40+30) = 70
similarly, for tangent 2, we will get (30+40) = 70 -----2

Hence, by 1&2, we will get 140+ 105Pi
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Re: Red ribbon string is put around two circular disks as shown above. If [#permalink] Red ribbon string is put around two circular disks as shown above. If   06 May 2020, 10:49
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Bunuel wrote:

Red ribbon string is put around two circular disks as shown above. If the radii of the two disks are 40 cm and 30 cm, respectively, what is the total length of the string?


A. \(140\) cm

B. \(140 + \frac{165π}{2}\) cm

C. \(140 + 105π\) cm

D. \(140 + 120π\) cm

E. \(140 + 165π\) cm

Show Spoiler
Attachment:
Untitled.png


If we break down the lengths we have: 2 arcs of 270 degrees. 2 lengths of 40, and 2 lengths of 30.

The straight lengths add up to 140 which doesn't help us determine the answer at all. Thus focus on finding the length of arcs.

Since it is 270 degree we find \(\frac{3}{4}\) of both circumferences. That would be \(2*\pi*(40+30) *\frac{3}{4} = 105*\pi\),

Hence we choose C.

Ans: C
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Re: Red ribbon string is put around two circular disks as shown above. If [#permalink] Red ribbon string is put around two circular disks as shown above. If   09 May 2020, 06:07
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Bunuel wrote:

Red ribbon string is put around two circular disks as shown above. If the radii of the two disks are 40 cm and 30 cm, respectively, what is the total length of the string?


A. \(140\) cm

B. \(140 + \frac{165π}{2}\) cm

C. \(140 + 105π\) cm

D. \(140 + 120π\) cm

E. \(140 + 165π\) cm



The length of the ribbon of each circle is 3/4 the circumference of the circle plus 2 times its radius.

Thus, the length of the ribbon of the 40-cm radius circle is ¾(2π x 40) + 2(40) = 60π + 80 cm. Similarly, the length of the ribbon of the 30-cm radius circle is ¾(2π x 30) + 2(30) = 45π + 60 cm. Thus the total length of the ribbon is 105π + 140 cm.

Answer: C
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Re: Red ribbon string is put around two circular disks as shown above. If [#permalink]
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